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Choked Flow For Vapor PSV Relief 3

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RJB32482

Chemical
Jan 19, 2005
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Here are the conditions of the relief scenario I am looking at:

Case: Fire
Set Pressure: 120 psig
Relief Flow: 350 SCFM
Relief Material: Air

Now there is about 8 feet of piping in between the outlet of the relief and its discharge into atmosphere. To find if the flow is choked, how would I calculate the pressure at the outlet of the relief device? Would it just be atmospheric pressure or atmospheric pressure+(pressure drop through piping at relief conditions). Then using inlet pressure and specific heatt ratios, I could find if the outlet of the relief device is choked or not.


Thanks.
 
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Find the resistance coefficient (Keq) of your discharge system. Using this value and the ratio of specific heats of air to deterime the pressure ratio. Crane technical paper No. 410 has charts on this. If your pressure ratio is greater than the critical pressure ratio (delta P)/(inlet P) then your flow is choked.
 
Thanks for the response. When doing this calc, do I have to take account any pressure drop in the valve itself? If so, how would I calculate the pressure drop in a PSV?

Thanks.
 
Yes. You must account for the pressure drop in the relief valve. You can probably get a Cv value for the valve from the manufacturer. Convert this to a K Value per the relationships in the Crane paper. Add to this the K values for the pipe lengths, fittings, entrances, exits, etc. Determine the pressure ratio as mentioned above. Use the pressure ratio to determine the flow rate through the system. Use the flow rate to determine the pressure drop across the relief valve.
 
Are you are referring to the charts for Net Expansion Factor Y for Compressible Flow Through Pipe to a Larger Flow Area? These charts were derived for a single diameter pipe. Combining a K for a PSV (the flow nozzle ~ 0.2") and a K for an outlet pipe (d ~ 2"), even if both K's are based on the smaller diameter, is not practical. The PSV K will overwhelm the numbers and make it difficult to get good numbers for the tailpipe.

Good luck,
Latexman
 
RJB32482,

Let's backup to your original post. The flow through a relief system can "choke" at several different points. Choked flow is also referred to as critical flow or sonic flow when the velocity of the fluid reaches its limiting velocity which is the velocity of sound in the flowing fluid at that location.

Typically, flow through a relief valve nozzle will experience critical flow unless the relief valve set pressure is very low (say about 15 psig) or you have substantial backpressure. You can use the methods in API RP-520 Part I to determine when you have critical or subcritical flow through your relief valve nozzle. When discharging to atmosphere, the backpressure (absolute) will be the sum of the pressure drop through the piping + atmospheric pressure. Compare your backpressure (absolute) to the critical flow pressure as determined in API to determine if you have critical or subcritical flow through your relief valve in order to select the proper relief valve sizing equations.

You can also experience critical (choked) flow in the relief valve outlet piping and typically this can occur at the end of the pipe where the fluid exits to atmosphere or at a point where the piping diameter changes such as at an expansion fitting. You can refer to the methods in API RP-521 to determine whether you experience critical flow in your outlet piping. Critical flow at points in the outlet piping will produce a pressure discontinuity and you can have a substantial pressure drop over small distances of travel and again it is related to the limiting velocity at that location. Any pressure discontinuities also need to be included as being part of the pressure drop through your piping.

I would add that there is some difference in the equations presented in 3rd (1990) and 4th (1997) editions of API RP-521. The method API RP-521 presents is based on an isothermal flow assumption. In the 3rd edition the equation for mach number is the isothermal mach number which I believe is the correct one to use for the isothermal assumption. The 4th edition gives the equation for mach number as the adiabatic mach number which differs only by inclusion of the specific heat ratio. I don't think the difference is great enough to be of significant concern but I think the next edition (5th) of API 521 is going back to showing the isothermal mach number equation.
 
Yes. Even with the limitations it is the best method that I am aware. The method would be to find the Keq of the valve best on the inside diameter of the 2" pipe and add that to the pipe lengths specified. See page 2-8 of Crane.


Seems like a very low beta for a relief nozzle. If this is this case, then the method simply indicates that the tail pipe has very little contribution to the back pressure.
 
Not only do I disagree with treating a PSV, which I treat as an isentropic nozzle, like a piece of pipe, but this straightforward PSV can be sized without combining the PSV with the outlet (or the inlet) into a K value. In fact, compressible flow methods do not have to be used on the inlet or outlet dP.

With a set pressure of 120 psi, we know the flow will be sonic at the flow nozzle exit in the PSV. 350 scfm sets the mass flow rate of the sizing basis (1604 lb/hr). If you pick a make, model, and flow nozzle size of PSV, this plus knowing Cp/Cv = 1.4 determines r[sub]c[/sub]. With the PSV fixed now, the actual peak flow rate through the PSV alone can be calculated and compared to the sizing basis. Ignore the inlet and outlet. Just calculate the flow through a wide open PSV with air at 120 x 1.21 psig.

Knowing the lengths and fittings, calculate the pipe diameter for the inlet (dP < 3% x 120 psig) and the outlet (dP < 10% x 120 psig) at the actual peak flow rate. Incompressible methods can be used here since dP < 10% of P[sub]1[/sub].

At this point, I believe Code has been satisfied and good engineering practice has been used. The PSV is big enough. Inlet dP < 3% of set pressure. Outlet dP < 10% of set pressure.

Good luck,
Latexman
 
By "Incompressible methods can be used here since dP < 10% of P1" I meant to say incompressible methods can be used here if we choose a diameter so dP < 10% of P1 (or P2),


Good luck,
Latexman
 
You are correct if the discharge pipe offers little resistance compared to the relief valve, but can you always assume that to be the case? I don't think so. You also cannot always assume the flow is incomopressible in the tail pipe. What if the relief valve had a high beta (like 0.8) and the discharge pipe long enough to have comparable resistance to the relief valve?
 
I agree you cannot always assume the flow is incompressible in the tail pipe. Good engineering practice suggests that when dP/P1 is greater than or equal to 40%, compressible flow methods must be used.

In a case where the tailpipe offers more resistance than this particular case, you can still keep the PSV and tailpipe flow calculation separate, but as the built-up backpressure increases you have to select a bellows PSV or pilot operated PSV that can handle the increased back-pressure. As mentioned above, the correct dP method for the tailpipe must be used.


Good luck,
Latexman
 
Fellow Tippers,

I strongly suggest you review the references in API RP-521 when comes to the design of relief valve discharge piping. I believe it offers a very reasonable and practical approach and I see no reason to re-invent the wheel on this subject.

As quoted from API RP-521
"Vapor flow in relief discharge piping is characterized by rapid changes in density and velocity; consequently, the flow should be rated as compressible."

Latexman, I believe you have mistakenly associated P1 with Pset in your evaluation of the use of an incompressible calculation for the outlet piping.

For Pset = 120 psig, and in the case of a conventional type relief valve, the allowable backpressure is 10% of Pset then Pb = 12 psig. If that backpressure is a result of builtup backpressure then dP = 12 psi.

For Pb = 12 psig or 26.7 psia this is in effect P1 of the outlet piping. In that case,
12 psi/26.7 psia = 0.45
which is well outside the limit of 0.1 in order to use the incompressible calculation approach with either the upstream or downstream denisty.

This is also beyond the limit of 0.4 where the incompressible approach is used with the average density between upstream and downstream conditions.

 
EGT01,

I can see how you interpreted it that way. I should have explained the nomenclature to avert this, instead of writing in "Crane TP410-ese", or separated the sentence they are in with more explanantion.

It was clear in my mind that I was talking about two separate conditions that each have a different 10% guideline. I put the Code requirement for the tailpipe in parentheses, like (dP < 10% x 120 psig), and said "incompressible methods can be used here if we choose a diameter so dP < 10% of P1". 120 psig is P[sub]set[/sub], P1 is the beginning of the tailpipe, and P2 is the end of the tailpipe. Nowhere did I say P1 = 120 psig, but the two similar looking guidelines are confusing in this case.


Good luck,
Latexman
 
Latexman,

Actually, I did not see your post at 12:09 before my post at 12:29 so your previous post helped to clear up that confusion.

Also, I definitely agree with you about keeping the relief valve calculation separate from the piping calculations.
 
EGT01,

Your earlier comment on API RP 521 Rev 4 isothermal flow equation is very interesting. I did a check of equation 21 in the course of some recent (subsea) flare header work, against Crane (eq 1-6) and GPSA (Eqn 17-15); after some algebra shuffling API seems to be short by Z/k on the right hand side. This isn't a trivial difference. I found it underestimating the header equivalent length by about 25% (Z~1, k~1.3) compared to the other two methods. Have you any feel for how conservative the isothermal assumption is for low pressure relief systems operating, say ~ Mach 0.5?
 
rbcoulter,

It's probably a bit late in the day, but for the record the choked flow check for the short tailpipe you discribed is very straightforward. We assume the exit loss at the end of the tailpipe equals 1 velocity head which conveniently makes the flare tip (flowing) pressure atmospheric. Assuming the boiling liquid in your fire case is water, the relieving temperature is around 357F (b.p. @ 1.1 x Pset). This gives a tip velocity of 446 fps or Mach 0.32 for AIR (steam gives lower Mach No.). In pipes of constant cross section (subsonic) Mach No.s increase with distance, therefore upstream Mach numbers are less than 0.32 and there is no choked condition.

Finally, providing the line loss calcs described by latexman and EGT01 pass code requirement of dP<0.1 x Pset, everything has checked out ok.

Hope this helps
 
sethoflagos,

I am not certain which case you are referencing. This thread refers back to RJB32482's scenario. I am not certain how you arrived at subsonic flow for the case you proposed. Maybe you can elaborate.
 
Oops, You're right of course, it was RJB32482's posting.

Anyway for the original case, of 350 scfm exiting the tailpipe tip at atmospheric from (I guessed Sch 80) 2" pipe, the velocity is subsonic (446 fps vs sonic velocity of ~1400 fps @ 357 deg F). Upstream, so long as there's no change in cross-section, the density is higher so the velocity is less. ie the exit velocity of 0.32 Mach at the tailpipe tip is the maximum Mach value.

Note that the only parameters here are exit pressure, mass flow and pipe geometry - the tailpipe hydraulics are independent of PRV set pressure.


However, providing downstream pressures are less than about half Pset, the PRV discharge nozzle is running in an altogether different regime that is dependent only on PRV set pressure and throat size - the PRV throat hydraulics are independent of exit pressure.

In this regime, flow through the PRV throat is sonic at a little below Pset. Exiting the throat, the gas accelerates and depressurises extremely rapidly in a free expansion jet that may achieve Mach 3 or so. Being supersonic, no downstream pressure information can travel against the flow which is why this zone is independent of exit pressure.

Eventually the two competing regimes must join but this requires the jet being able to match both the downstream pressure and velocity. It can only do this by converting a velocity drop to a pressure increase (conservation of momentum) in an adiabatic recompression. Because there is so much velocity to lose it has to first expand to substially below atmospheric pressure to make room for the positive step in pressure. When it reaches a point where everything balances the jet can jump instantaneously to the downstream subsonic regime creating a pressure discontinuity and a front of pressure shockwaves. Typically, this all happens within the PRV discharge nozzle or immediately downstream, so for all the noise it makes, it actually has little impact on tailpipe hydraulics.





 
I understand your point now. Of course, the choice of the 2" discharge pipe is arbitrary since it wasn't mentioned in the original post. Assuming that that is indeed the case then you are correct that the flow will most likely be choked at the relief valve orifice or nozzle because the delta p to upstream pressure ratio is quite high. You seem to be speculating about what occurs downstream of the nozzle. I don't really know. I have read that supersonic velocity can occur but my understanding is that you need a certain shape expansion nozzle to do this. Most likely, the energy is lost to shock waves, turbulence, etc. without achieving sonic velocity. I am basically paraphrasing what I have read in my Thermodynamics textbook from college which I keep on my computer desk.
 
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